Raising the bar Properties of root subspaces: geometry that will lead to Jordan form

Jul 19

Properties of root subspaces

Properties of root subspaces

Let $A$ be a square matrix and let $\lambda \in \sigma (A)$ be its eigenvalue. As we know, the nonzero elements of the null space $N(A-\lambda I)=\{x:(A-\lambda I)x=0\}$ are the corresponding eigenvectors. This definition is generalized as follows.

Definition 1. The subspaces $N_{\lambda }^{(k)}=N((A-\lambda I)^{k}),$ $k=1,2,...$ are called root subspaces of $A$ corresponding to $\lambda .$

Exercise 1. a) Root subspaces are increasing:

(1) $N_{\lambda }^{(k)}\subset N_{\lambda }^{(k+1)}$ for all $k\geq 1$

and b) there is such $p\leq n$ that all inclusions (1) are strict for $k<p$ and

(2) $N_{\lambda }^{(p)}=N_{\lambda }^{(p+1)}=...$

Proof. a) If $x\in N_{\lambda }^{(k)}$ for some $k,$ then $(A-\lambda I)^{k+1}x=(A-\lambda I)(A-\lambda I)^{k}x=0,$ which shows that $x\in N_{\lambda }^{(k+1)}.$

b) (1) implies $\dim N_{\lambda }^{(k)}\leq \dim N_{\lambda }^{(k+1)}.$ Since all root subspaces are contained in $C^{n},$ there are $k$ such that $N_{\lambda }^{(k)}=N_{\lambda }^{(k+1)}.$ Let $p$ be the smallest such $k.$ Then all inclusions (1) are strict for $k<p.$

Suppose $N_{\lambda}^{(k+1)}\setminus N_{\lambda }^{(k)}\neq \varnothing$ for some $k\ge p.$ Then there exists $x\in N_{\lambda }^{(k+1)}$ such that $x\notin N_{\lambda}^{(k)}$ , that is, $(A-\lambda I)^{k+1}x=0,$ $(A-\lambda I)^{k}x\neq 0.$ Put $y=(A-\lambda I)^{k-p}x.$ Then $(A-\lambda I)^{p+1}y=(A-\lambda I)^{k+1}x=0,$ $(A-\lambda I)^{p}y=(A-\lambda I)^{k}x\notin 0.$ This means
that $y\in N_{\lambda }^{(p+1)}\setminus N_{\lambda }^{(p)}$ which contradicts the definition of $p.$

Definition 2. Property (2) can be called stabilization. The number $p$ from (2) is called a height of the eigenvalue $\lambda$ .

Exercise 2. Let $\lambda \in \sigma (A)$ and let $p$ be the number from Exercise 1. Then

(3) $C^{n}=N_{\lambda }^{(p)}\dotplus \text{Img}[(A-\lambda I)^{p}].$

Proof. By the rank-nullity theorem applied to $(A-\lambda I)^{p}$ we have $n=\dim N_{\lambda }^{(p)}+\dim \text{Img}[(A-\lambda I)^{p}].$ By Exercise 3, to prove (3) it is sufficient to establish that $L\equiv N_{\lambda}^{(p)}\cap \text{Img}[(A-\lambda I)^{p}]=\{0\}.$ Let's assume that $L$ contains a nonzero vector $x.$ Then we have $x=(A-\lambda I)^{p}y$ for some $y.$ We obtain two facts:

$(A-\lambda I)^{p}y\neq 0$ $\Longrightarrow y\notin N_{\lambda }^{(p)},$

$(A-\lambda I)^{2p}y=(A-\lambda I)^{p}(A-\lambda I)^{p}y=(A-\lambda I)^{p}x=0\Longrightarrow y\in N_{\lambda }^{(2p)}.$

It follows that $y$ is a nonzero element of $N_{\lambda }^{(2p)}\setminus N_{\lambda }^{(p)}.$ This contradicts (2). Hence, the assumption $L\neq \{0\}$ is wrong, and (3) follows.

Exercise 3. Both subspaces at the right of (3) are invariant with respect to $A.$

Proof. If $x\in N_{\lambda }^{(p)},$ then by commutativity of $A$ and $A-\lambda I$ we have $(A-\lambda I)^{p}Ax=A(A-\lambda I)^{p}x=0,$ so $Ax\in N_{\lambda }^{(p)}.$

Suppose $x\in \text{Img}[(A-\lambda I)^{p}],$ so that $x=(A-\lambda I)^{p}y$ for some $y.$ Then $Ax=(A-\lambda I)^{p}Ay\in \text{Img}[(A-\lambda I)^{p}].$

Exercise 3 means that, for the purpose of further analyzing $A,$ we can consider its restrictions onto $N_{\lambda }^{(p)}$ and $\text{Img}[(A-\lambda I)^{p}].$

Exercise 4. The restriction of $A$ onto $N_{\lambda }^{(p)}$ does not have eigenvalues other than $\lambda .$

Proof. Suppose $Ax=\mu x,$ $x\neq 0,$ for some $\mu .$ Since $x\in N_{\lambda }^{(p)},$ we have $(A-\lambda I)^{p}x=0.$ Then $(A-\lambda I)x=(\mu -\lambda )x$ and $0=(A-\lambda I)^{p}x=(\mu -\lambda )^{p}x$ . This implies $\mu =\lambda .$

Exercise 5. The restriction of $A$ onto $\text{Img}[(A-\lambda I)^{p}]$ does not have $\lambda$ as an eigenvalue (so that $A-\lambda I$ is invertible).

Proof. Suppose $x\in \text{Img}[(A-\lambda I)^{p}]$ and $Ax=\lambda x,$ $x\neq 0.$ Then $x=(A-\lambda I)^{p}y$ for some $y\neq 0$ and $0=(A-\lambda I)x=(A-\lambda I)^{p+1}y.$ By Exercise 1 $y\in N_{\lambda }^{(p+1)}=N_{\lambda }^{(p)}$ and $x=(A-\lambda I)^{p}y=0.$ This contradicts the choice of $x.$